What is the general solution for cos(2θ)+sin^2(θ)=0,0<= θ<2pi ?

The general solution for cos(2θ)+sin^2(θ)=0,0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2

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cos(2θ)+sin^2(θ)=0,0<= θ<2pi

cos (2 θ) + sin^{2} (θ) = 0, 0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

cos (2 θ) + sin^{2} (θ) = 0, 0 \leq θ < 2 π

Rewrite using trig identities

cos (2 θ) + sin^{2} (θ)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

- sin^{2} (x) = cos (2 x) - cos^{2} (x)

= cos (2 θ) - (cos (2 θ) - cos^{2} (θ))

Simplify

cos (2 θ) - (cos (2 θ) - cos^{2} (θ)) : cos^{2} (θ)

cos (2 θ) - (cos (2 θ) - cos^{2} (θ))

- (cos (2 θ) - cos^{2} (θ)) : - cos (2 θ) + cos^{2} (θ)

- (cos (2 θ) - cos^{2} (θ))

Distribute parentheses

= - (cos (2 θ)) - (- cos^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (2 θ) + cos^{2} (θ)

= cos (2 θ) - cos (2 θ) + cos^{2} (θ)

Add similar elements:

cos (2 θ) - cos (2 θ) = 0

= cos^{2} (θ)

= cos^{2} (θ)

cos^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

tan^{2} (x) + tan (x) = 0

cos (x) - 1 = 0

cos^{2} (x) = 0

sec (x) csc (x) = 2 csc (x)

3 csc^{2} (x) - 4 = 0

What is the general solution for cos(2θ)+sin^2(θ)=0,0<= θ<2pi ?
The general solution for cos(2θ)+sin^2(θ)=0,0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2